Question about elements in a cartesian closed category.

Question

I was recently studying the basics of topoi for a class and had some questions about cartesian closed categories, more specifically about elements and exponentials (from here on out, $\simeq$ means an isomorphism in the corresponding category).

1. Is is true that, if $1$ is a terminal object in $\mathcal{C}$, then $\mathcal{C}(1,X)\simeq\mathcal{C}(1,Y)$ implies $X\simeq Y$? I know that's true in $\mathtt{Set}$ (and in any category in which we can actually regard $X^Y$ as $\mathcal{C}(Y,X)$ not only by $\mathcal{C}(1,X^Y)\simeq\mathcal{C}(Y,X)$, but in a more "concrete" sense);
2. (I already know how to do this if the above item is true) Is it true that if $F$ and $G$ are both $\mathcal{C}$-endofunctors such that $F\dashv G$, then we not only have $$\mathcal{C}(FX,Y)\simeq\mathcal{C}(X,GY)$$ but also $$Y^{FX}\simeq GY^X?$$

I'm really interested in these results and greatly appreciate any help!

Answer 1

1. No. Toposes in which that statement is true are called well-pointed and they are relatively rare things. Cartesian closure is therefore certainly not enough to induce it. To connect what you said more closely to well-pointedness, note that your statement is essentially that $\mathcal{C}(1,-)$ is conservative. A conservative functor is faithful if it preserves equalizers and $\mathcal{C}$ has them. $\mathcal{C}(1,-)$, like all covariant hom-functors, preserves all limits including equalizers. Toposes, by definition, have equalizers.

2. This condition implies that $F$ is a strong functor (which it always will be in $\mathbf{Set}$). In particular, it implies $F(A\times X)\cong A\times FX$ natural in $A$ and $X$. A strong functor is more or less one where the functorial action can be internalized meaning instead of just having a family of functions $\mathcal{C}(A,B)\to\mathcal{C}(FA,FB)$ we have a natural transformation $[A,B]\to[FA,FB]$ where $[A,B]$ is the internal hom of $\mathcal{C}$ which, with respect to the cartesian monoidal structure, is $B^A$. At any rate, functors don't need to be strong and it should already have been a warning sign that this isomorphism required $F$ and $G$ to be endofunctors.